proof of a finite sum involving a binomial coefficient and a variable. I found that the following equation holds for integers $l$, $k$, and any $x \neq 0,1$,
$$\tag{1}
\sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
   k  \\
   l  \\
\end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{{1 + l}}{x} + k - l} \right)\left( {\frac{l}{x} + 1 + k - l} \right)}} = \frac{x\left( {1 - x} \right)^{k}}{{k + 1}}
$$
both in numerically by Matlab and analytically by Mathematica. 
So I think there is a reference proving the equation. I have searched equaitons in Wolfram, Wiki, and some tables of series,  But I couldn't find any related one.
Actually there were some equations looks like this
$$\tag{2}
\sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
   k  \\
   l  \\
\end{array}} \right)f(k,l,x) = g(k,x),
$$
but no help.
Also I tried in this way: break the equation into two terms like this
$$\tag{3}
\sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
   k  \\
   l  \\
\end{array}} \right)\frac{{x^{l}  }}{{\left( {\frac{{1 + l}}{x} + k - l} \right)}} + \sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
   k  \\
   l  \\
\end{array}} \right)\frac{{x^{l}  }}{{\left( {\frac{l}{x} + 1 + k - l} \right)}},
$$
and ran it in Mathematica. But they result in Gauss hypergeometric functions, $_2 F_1 (-k,*,*,x)$, with some coefficients, respectively.
Also $\times2$, I tried to prove it by myself showing the equation holds for $k=0$ and any $x \neq0,1$, then it holds as well when $k+1$ by using the eq (1). but I couldn't...beacuse the $k+1$ case becomes a totally different equation compared to eq (1)...... lol 
How can I prove it or find a proof?
 A: We have, $$\frac{1}{\left(\frac lx + k - l+1 \right) \left(\frac{l+1}{x} + k-l \right)} = \frac{x}{1-x} \left(\frac{1}{\frac lx + k - l+1} - \frac{1}{\frac{l+1}{x} + k-l} \right)$$
Therefore, $$\sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\left(\frac lx + k - l+1 \right) \left(\frac{l+1}{x} + k-l \right)} = \frac{x}{1-x} \left(\sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac lx + k - l+1} - \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac{l+1}{x} + k-l} \right)$$
Call the term inside the bracket as $S$. So, we only need to show that $S = \displaystyle \frac{(1-x)^{k+1}}{k+1}$.
Now, $\displaystyle \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac lx + k - l+1} = \frac 1{k+1} + \sum_{l=0}^{k-1} (-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} $, and
$\displaystyle \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac {l+1}x + k - l} = \frac {(-1)^k x^{k+1}}{k+1} + \sum_{l=0}^{k-1} (-1)^{l} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l} $
Thus, on subtracting, we get that
$$S = \frac{1}{k+1} + \frac {(-x)^{k+1}}{k+1} + \sum_{l=0}^{k-1} \left((-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} + (-1)^{l+1} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l}\right)$$
Now, since $\binom{k}{l+1} = \frac{k-l}{l+1} \binom{k}{l}$, therefore,
$$(-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} + (-1)^{l+1} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l} = (-1)^{l+1} \binom{k}{l} \frac{x^{l+1}}{l+1 +(k-l)x} \left(\frac{k-l}{l+1} x + 1\right) = (-1)^{l+1} \binom{k}{l} \frac{x^{l+1}}{l+1} = (-1)^{l+1} \binom{k+1}{l+1} \frac{x^{l+1}}{k+1}$$
Thus, $\displaystyle S = \sum_{l=0}^{k+1} \binom{k+1}{l} \frac{x^l}{k+1} = \frac{(1-x)^{k+1}}{k+1}$, which was what we needed.
